Problem: Nicholas sent a chain letter to his friends, asking them to forward the letter to more friends. Every $12$ weeks, the number of people who receive the email increases by an additional $99\%$, and can be modeled by a function, $P$, which depends on the amount of time, $t$ (in weeks). Nicholas initially sent the chain letter to $50$ friends. Write a function that models the number of people who receive the email $t$ weeks since Nicholas initially sent the chain letter. $P(t) = $
Answer: The strategy We can model the situation with an exponential function of the general form A ⋅ B f ( t ) A\cdot B\^{ f(t)}, where $A$ is the initial quantity, $B$ is a factor by which the quantity is multiplied over constant time intervals, and $f(t)$ is an expression in terms of $t$ that determines those time intervals. Let's use the given information to determine $A$, $B$, and $f(t)$. Understanding what's given We are given that the initial number of people who receive the email is $50$, and every $12$ weeks, the number of people who receive the email increases by an additional $99\%$. Note that increasing by $99\%$ is the same as being multiplied by $1.99$. [Why?] This means that the initial quantity is $A=50$ and the factor is $B=1.99$. We need to find $f(t)$ based on the fact that the quantity is multiplied by $1.99$ every $12$ weeks. Finding the expression in the exponent We know that the number of people who receive the email is multiplied by $1.99$ every $12$ weeks. This means that each time $t$ increases by $12$, $f(t)$ increases by $1$. Therefore, $f(t)$ is a linear function whose slope is $\dfrac{1}{12}$. When Nicholas initially sent the e-mail, the number of people who have received the e-mail has not begun to increase yet. So $P(0) = 50$, which means that $f(0)=0$. [Why?] Therefore, $f(t)$ must be $\dfrac{t}{12}$. Summary We found that the following function models the number of people who receive the email $t$ weeks since Nicholas initially sent the chain letter. P ( t ) = 50 ⋅ ( 1.99 ) t 12 P(t)=50\cdot (1.99)\^{ \frac{t}{12}}